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798
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
(c)
The equality obtained in part (b) implies that
λ
is the rate of change in the maximum value of
f
(
x
,
y
)
, subject to the
constraint
g
(
x
,
y
)
=
c
, with respect to
c
.
41.
Let
B
>
0. Show that the maximum of
f
(
x
1
,... ,
x
n
)
=
x
1
x
2
···
x
n
subject to the constraints
x
1
+···+
x
n
=
B
and
x
j
≥
0for
j
=
1
n
occurs for
x
1
=···=
x
n
=
B
/
n
. Use this to conclude that
(
a
1
a
2
a
n
)
1
/
n
≤
a
1
a
n
n
for all positive numbers
a
1
a
n
.
SOLUTION
We Frst notice that the constraints
x
1
x
n
=
B
and
x
j
≥
j
=
1
n
deFne a closed and
bounded set in the
n
th dimensional space, hence
f
(continuous, as a polynomial) has extreme values on this set. The
minimum value zero occurs where one of the coordinates is zero (for example, for
n
=
2 the constraint
x
1
+
x
2
=
B
,
x
1
≥
0,
x
2
≥
0 is a triangle in the Frst quadrant). We need to maximize the function
f
(
x
1
x
n
)
=
x
1
x
2
x
n
subject to the constraints
g
(
x
1
x
n
)
=
x
1
x
n
−
B
=
0,
x
j
≥
0,
j
=
1
n
.
Step 1.
Write out the Lagrange Equations. The gradient vectors are
∇
f
=

x
2
x
3
x
n
,
x
1
x
3
x
n
x
1
x
2
x
n
−
1
®
∇
g
= h
1
,
1
1
i
The Lagrange Condition
∇
f
=
∇
g
yields the following equations:
x
2
x
3
x
n
=
x
1
x
3
x
n
=
x
1
x
2
x
n
−
1
=
Step 2.
Solving for
x
1
,
x
2
x
n
using the constraint. The Lagrange equations imply the following equations:
x
2
x
3
x
n
=
x
1
x
2
x
n
−
1
x
1
x
3
x
n
=
x
1
x
2
x
n
−
1
x
1
x
2
x
4
x
n
=
x
1
x
2
x
n
−
1
.
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 Winter '08
 GANGliu
 Rate Of Change

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